top space / manifold structure on points of alg varieties

Stream: maths

[Kevin Buzzard]

If $X$ is an affine algebraic variety (or just an affine scheme of finite type) over a field $K$, and if $\mathcal{O}_X(X)=A$, and if $R$ is a commutative $K$-algebra with a topology making $R$ into a topological ring, then I imagine that the $R$-points $X(R)$ (i.e. the $K$-algebra homomorphisms from $A$ to $R$) inherit a canonical topology; for example one can write down a closed immersion from $X$ into some large affine space $\mathbb{A}^N$ and this identifies $X(R)$ with a subspace of $R^N$, so we can give it the subspace topology. My instinct is that this topology is independent of the choice of closed immersion but I just spent a few minutes writing down what I thought would be the proof (given two embeddings into $R^N$ and $R^M$ consider the product embedding and then look at projections) and couldn't immediately finish it off. What trick am I missing?

Now say $X$ is also smooth, and $K$ is a nontrivially normed field. Then do the same sort of ideas give $X(K)$ the structure of a manifold over $K$, again independent of all choices? Smoothness is some criterion about differentials behaving which will be enough to make things like the implicit function theorem kick in, and then I get stuck at the same point as above.

The internet seems to be full of people saying that all of this is obvious. How hard will it be to formalise? I need it for FLT :-)

[Adam Topaz]

oh you don't need to embed anything. Just take the coarsest topology making the algebraic functions continuous.

[Adam Topaz]

you could formalize the topology as some infimum of some induced topologies, etc.

[Adam Topaz]

as for smoothness and the manifold structure, that's of course a much bigger deal.

[Kevin Buzzard]

I think that the coarsest topology making everything continuous is embedding it into affine A-space :-) So then the question becomes: is it true that this is the subspace topology for a closed embedding? Concretely, if my affine variety is GL_n and the topological ring is the reals, how do I prove that the topology you're suggesting agrees with the two topologies I've seen in the literature, namely those obtained by embedding into R^{n^2+1} via mapping M to (M,1/det(M)) and by embedding into R^{2n^2} via (M,1/M)?

[Adam Topaz]

but the point is that you don't even need to assume that it's affine.

[Adam Topaz]

If $X$ is any variety over $K$, and $U$ is Zariski open, then $f \in \mathcal{O}(U)$ induces $U(K) \to K$. Now take the induced topology with respect to this map on $U(K)$ and take the infimum as all $f$ vary. Now look at the coinduced topology on $X(K)$, then take the supermum of all such as $U$ varies.

[Adam Topaz]

I think that should work.

[Adam Topaz]

I think it's also possible to characterize this as the coarsest family of topologies on $X(K)$ as $X$ vary over all $K$-varieties for which all $f \in \mathcal{O}(X)$ induce continuous map $X(K) \to K$ and for which $X \to Y$ induce continuous maps $X(K) \to Y(K)$ for any morphism of varieties

[Adam Topaz]

I guess the condition on $X(K) \to K$ is superfluous since you can just look at morphisms $X \to \mathbb{A}^1$ (once you identify the topology on $\mathbb{A}^1(K) = K$ with the strong one) :)

[Kevin Buzzard]

Right but you're not answering the question. I don't doubt that your idea is a good idea but what I'm saying is that in the special case of GL_n(R) this is not how the topology is defined in the literature.

[Kevin Buzzard]

Is my guess that you can use any closed embedding (rather than all of them, which is what you're doing) not actually correct?

[Kevin Buzzard]

And does the same trick work for the manifold structure? I guess I can just open GAGA and see what Serre does

[Adam Topaz]

Oh sorry, yeah I didn't read :)

[Adam Topaz]

... I've been working on this #11964

[Mitchell Lee]

If $X$ is affine, I believe the topology you describe is indeed independent of the choice of closed immersion.

Let $X \to \mathbb{A}^N$ be a closed immersion and let $f_1, \ldots, f_N \in \mathcal{O}(X)$ be the $N$ coordinate functions. The topology you describe is the coarsest topology such that for all $i$, the evaluation $X(R) \to R$ of $f_i$ is continuous. Since $f_1, \ldots, f_N$ generate $\mathcal{O}(X)$, this is equivalent to the statement that the evaluation $X(R) \to R$ of $f$ is continuous for all $f \in \mathcal{O}(X)$.

[Adam Topaz]

Right. You would have to show that polynomial functions are continuous for that proof. Do we have that?

[Mitchell Lee]

docs#MvPolynomial.continuous_eval

[Mitchell Lee]

In fact, I suspect that if we put a topology on $X(R)$ using the method Adam Topaz describes, then:

1. If $X \to Y$ is a closed immersion, then $X(R) \to Y(R)$ is a closed embedding.
2. $(X \times Y)(R) = X(R) \times Y(R)$ has the product topology.

[Mitchell Lee]

And then, one more fact is needed to deduce the theorem you want, which is that the topology on $\mathbb{A}^1(R) = R$ is the same as the original one that you put on $R$.

[Adam Topaz]

I’m not sure if it’s true in that level of generality. And in particular some assumptions on the ring are probably required

[Adam Topaz]

Cf. https://arxiv.org/pdf/2009.02319.pdf

[Adam Topaz]

Well Fact 4.7 shows that it’s ok in the case of a field

[Adam Topaz]

(Which comes from the red book)

[Adam Topaz]

Ok, now I’m curious to see in what level of generality this holds :)

[Adam Topaz]

What if we put a topology on $\mathcal{O}(S)$ for some scheme S. will it hold for the X(S)’s for X separated and of finite type over S?

[Kevin Buzzard]

I went for topological rings because I wanted to apply this for the finite adeles of a number field, which are hideously non-Noetherian.

[Adam Topaz]

@Kevin Buzzard I found these notes by Conrad which makes things pretty explicit: https://math.stanford.edu/~conrad/papers/adelictop.pdf

[Kevin Buzzard]

oh many thanks! I had considered asking Brian instead of asking here, but I've been badgering him about finite flat group schemes for a while now so thought I'd give him a break :-)

[Antoine Chambert-Loir]

I was going to give the same reference. It is not so trivial, though, as the case of adeles can suggest.

[Mitchell Lee]

For the manifold question, here is the "low-powered" way. Suppose you have two closed immersions $X \to \mathbb{A}^m$ and $X \to \mathbb{A}^n$. We wish to show that the corresponding closed immersions $f \colon X(K) \to K^m$ and $g \colon X(K) \to K^n$ induce the same analytic manifold structure on $X(K)$.

For any $x \in X(K)$, one can construct an analytic function $h \colon U \to K^n$, defined on a neighborhood $U$ of $f(x)$ in $K^m$, such that $h(f(y)) = g(y)$ for $y$ near $x$. (One can construct $h$ by composing $g$ with a function that is locally a left inverse of $f$. This exists by the implicit function theorem.) So the identity function $\operatorname{id}_{X(K)}$ is locally analytic at $x$, where the manifold structure on the domain is induced by $f$ and the manifold structure on the codomain is induced by $g$. You can do the same in the reverse direction.

[Mitchell Lee]

Hopefully I have not misunderstood what is going on here.

[Kevin Buzzard]

This makes me think that there should be an analogous low-level argument to show that any two closed immersions X -> A^n should induce the same topology on X(R).

[Mitchell Lee]

Now that I think about this argument, maybe it needs a little more work.

I am referring to the work required to prove that $h$ is actually analytic. I think you can't just say "it's analytic because it's the composition of two analytic functions", because that would be circular reasoning.

[Mitchell Lee]

Even if this works, I'm not convinced there's an analogous argument for the topology on $X(R)$. This seems to intrinsically rely on the implicit function theorem.

[Antoine Chambert-Loir]

Kevin Buzzard said:

This makes me think that there should be an analogous low-level argument to show that any two closed immersions X -> A^n should induce the same topology on X(R).

Take the diagonal embedding in A^m x A^n. Taking the induced topologies of the product topology of R gives you 3 topologies on X, say T1 T2 and T. The maps from (X,T) to (X,T1) and (X,T2) are continuous since they just forget about coordinates, and one needs to show that they are homeomorphisms. Now, by definition of a closed immersion (Nullstellensatz), the second embedding of X to A^n factors through a map A^m -> A^n. Consequently, (X,T1)->(X,T) is of the form (x1,...,xm) -> (x1,...,xm, f1(x1,..,xm),...,fn(x1,...,xm)) where f1,..., fn are polynomials, and it follows from the definition of a topological ring that they induce continuous functions R^m ->R.

[Mitchell Lee]

I didn't realize that you could just factor the map algebraically. That also gives a cleaner proof that the analytic structure is preserved.